Comparative Statics with Small Changes

Math Alert: Comparative Statics with Small Changes:

Let’s assume that the relevant “environmental” variable only affects the supply function. In this case we can write the demand function solely as a function of the own price: QD = D(p), but the supply function as: QS = S(p, a). At the initial equilibrium we have (by definition of equilibrium):

D(p) = S(p, a)

We know that as an environmental variable changed the equilibrium price will (eventually) change. So we may think of the price, not as a purely exogenous variable, but as an implicit function of the environmental variable a, or p = p(a). We therefore rewrite the equilibrium condition as:

D(p(a)) = S(p(a), a)

The total change in quantity demanded from one equilibrium to another is dD(p(a))/dp and the change in the equilibrium price due to a change in a is dp/da. The total change in quantity supplied is due both to the price change (which in turn depends on how much the price changes with each unit change in a) and by how much supply changes (directly) due to the change in a:

From the initial to the final equilibrium the total change in quantity demanded must be equal to the total change in quantity supplied:

We want to know by how much the equilibrium price changes due to a small increase in a, so we have to solve this equation for the term, dp/da:

The “Law of Demand” says that quantity demanded will decrease as the price increases; hence dD/dp < 0, if the supply function is upward sloping it follows that: ∂S/∂p > 0, which means that the denominator is negative, hence the sign of the dp/da has the opposite sign as the numerator (∂S/∂a ), if ∂S/∂a is negative dp/da is positive, and vice versa.

Example:

As an example, consider the following demand function for beer:

QD = 100 − 2. p + 0.05. ps − 0.25 . pc − 0.00125 . Y

QD is here expressed in a volume measure (liters or bottles or cans etc) per month and prices and income in money (kr, €, $, £ etc). The substitute good maybe wine and the complement good sausage. If p = 10kr/bottle, ps = 50kr/bottle, pc = 30kr/kilo and Y = 20000kr/month.

We can rewrite the equilibrium condition as:

70 − 2. p = 22 + 3. p − 2 . pm,

and solve for the price of beer as a function of the price of malt:

p = (48/5) + (2/5) . pm

The derivative of the equilibrium price with respect to a change in the price of malt is therefore:

dp/dpm = 2/5. The demand function is:

Q = D(p(pm)) = 70 − 2 . p(pm)

The change in quantity demanded due to a small change in the price of malt is therefore:

Note that this was exactly how much the price decreased by in the numerical example we looked at before.

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